Solve the pairs of linear equation by the elimination method and …

Question

Solve the pairs of linear equation by the elimination method and the substitution method:$\frac{x}{2} + \frac{{2y}}{3} = - 1\,and\,x - \frac{y}{3} = 3$

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Answer

By Elimination method
The given system of equation is
$\frac { x } { 2 } + \frac { 2 y } { 3 } = - 1...............(1)$
$x - \frac { y } { 3 } = 3....................(2)$
Multiplying equation $(2)$ by $2,$ we get
$2 x - \frac { 2 y } { 3 } = 6......................(3)$
Adding equation$(1)$ and equation $(2),$ we get
$\frac { 5 } { 2 } x = 5 \Rightarrow x = \frac { 5 \times 2 } { 5 } \Rightarrow \quad x = 2$
Substituting this value of $x$ in equation$(2),$ we get
$2 - \frac { y } { 3 } = 3 \Rightarrow \frac { y } { 3 } = 2 - 3 = - 1 \Rightarrow \quad y = - 3$
So, the solution of the given system of equation is
$x = 2, y = -3$
By substitution method
The given system of equation is
$\frac { x } { 2 } + \frac { 2 y } { 3 } = - 1...............(1)$
$x - \frac { y } { 3 } = 3....................(2)$
From equation $(2),$
$x = \frac { y } { 3 } + 3....................(3)$
Substituting this value of $x$ in $(1),$
$\frac { 1 } { 2 } \left( \frac { y } { 3 } + 3 \right) + \frac { 2 y } { 3 } = - 1$
$\Rightarrow \quad \frac { y } { 6 } + \frac { 3 } { 2 } + \frac { 2 y } { 3 } = - 1 \Rightarrow \quad \frac { 5 y } { 6 } = - 1 - \frac { 3 } { 2 }$
$\Rightarrow \quad \frac { 5 y } { 6 } - - \frac { 5 } { 2 } \Rightarrow y = - 3$
Substituting this value of y in equation $(3)$, we get
$x = - \frac { 3 } { 3 } + 3 = - 1 + 3 - 2$
So, the solution of the given system of equations is $x = 2, y = -3$
Verification: Substituting $x = 2, y = -3,$ we find that both the equation $(1)$ and $(2)$ are satisfied as shown below:
$\frac { x } { 2 } + \frac { 2 y } { 3 } = \frac { 2 } { 2 } + \frac { 2 ( - 3 ) } { 3 } = 1 - 2 = - 1$